$A$ charged $30\; \mu F$ capacitor is connected to a $27\; mH$ inductor. The initial charge on the capacitor is $6\; mC$. What is the total energy stored in the circuit initially? What is the total energy at a later time?

(A) The capacitance of the capacitor is $C = 30\; \mu F = 30 \times 10^{-6}\; F$.
The inductance of the inductor is $L = 27\; mH = 27 \times 10^{-3}\; H$.
The initial charge on the capacitor is $Q = 6\; mC = 6 \times 10^{-3}\; C$.
The total energy stored in the circuit is initially stored in the electric field of the capacitor,given by $E = \frac{1}{2} \frac{Q^2}{C}$.
Substituting the values: $E = \frac{1}{2} \times \frac{(6 \times 10^{-3})^2}{30 \times 10^{-6}} = \frac{1}{2} \times \frac{36 \times 10^{-6}}{30 \times 10^{-6}} = \frac{1}{2} \times 1.2 = 0.6\; J$.
Since the $LC$ circuit is assumed to be ideal (no resistance),the total energy is conserved and oscillates between the electric field of the capacitor and the magnetic field of the inductor. Therefore,the total energy at a later time remains $0.6\; J$.